Implement intersection of arrays (with duplicates & order)

Question

Accepted Answer

**Approach:** Use `Set` for O(1) lookup for unique, or `Map` counting for duplicates.

```javascript
function intersection(a, b) {
  const setB = new Set(b);
  return [...new Set(a.filter(x => setB.has(x)))];
}
function intersectionWithDup(a, b) {
  const count = new Map();
  for (const x of b) count.set(x, (count.get(x)||0)+1);
  const res = [];
  for (const x of a) { const c = count.get(x) || 0; if (c > 0) { res.push(x); count.set(x, c-1); } }
  return res;
}

// Demo
console.log(intersection([1,2,2,3], [2,3,4])); // [2,3]
console.log(intersectionWithDup([1,2,2,3], [2,2,4])); // [2,2]
console.log(intersection(['a','b','c'], ['b','c','d'])); // ['b','c']
console.log(intersection([], [1,2])); // []
function intersectionAll(...arrays) { return arrays.reduce((a,b) => intersection(a,b)); }
console.log(intersectionAll([1,2,3],[2,3,4],[3,4,5])); // [3]
```
Use `Set` intersection (ES2024 `a.intersection(b)`) when available.

Implement intersection of arrays (with duplicates & order)

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